The vanishing ideal $I_{K[x,y]}(A\!\times\!B)$ is generated by $I_{K[x]}(A) \cup I_{K[y]}(B)$?

Question

Let $K$ be a field, $x=(x_1,\ldots,x_m)$, $y=(y_1,\ldots,y_n)$, $A\!\subseteq\!\mathbb{A}^m_K$, $B\!\subseteq\!\mathbb{A}^n_K$. Does there hold $$I_{K[x,y]}(A\!\times\!B)=\langle\langle I_{K[x]}(A) \cup I_{K[y]}(B)\rangle\rangle?$$ Here $I_{K[x]}(A)=\{f\in K[x]; f(a)\!=\!0\text{ for all }a\!\in\!A\}$ is the vanishing ideal of set $A$, and $\langle\langle\ldots\rangle\rangle$ is the ideal, generated by $\ldots$. The inclusion $\supseteq$ is easy, but I don't see how to show $\subseteq$.

If this equality does not hold, how else then can I prove (knowing $$K[x,y]/\langle\langle\mathfrak{a},\mathfrak{b}\rangle\rangle \,\cong\, K[x]/\mathfrak{a}\otimes_KK[y]/\mathfrak{b}$$ from this post), that there is an isomorphism of $K$-algebras (coordinate rings) $$K[A\!\times\!B] \,\cong\, K[A]\otimes_K K[B]?$$

Answer 1

This is well-known and can be found in any some introductions to classical algebraic geometry, in the section about products. Anyway, here is my favorite proof. I assume that $k$ is algebraically closed (otherwise it is wrong).

We have $V(I(A \times B))=A \times B = (A \times \mathbb{A}^n) \cap (\mathbb{A}^m \times B) = V(I(A) \cup I(B))$, thus $I(A \times B) = \sqrt{I(A) + I(B)}$. We have to prove that $I(A) + I(B)$ is a radical ideal, or equivalently that $k[x,y]/(I(A)+I(B))=k[X]/I(A) \otimes k[Y]/I(B)$ is reduced (i.e. $0$ is the only nilpotent element).

Lemma. If $k$ is an algebraically closed field and $R,S$ are reduced $k$-algebras, then $R \otimes_k S$ is reduced.

Proof: A colimit argument shows that we may assume that $R$ is of finite type over $k$. Since $R$ is reduced, the intersection of all prime ideals is $0$, which equals the intersection of all maximal ideals since $R$ is jacobson. This gives an embedding $R \hookrightarrow \prod_{\mathfrak{m} \in \mathrm{Spm}(R)} R/\mathfrak{m}$, where $R/\mathfrak{m}=k$. This induces an embedding of $k$-algebras

$$R \otimes_k S \hookrightarrow (\prod_{\mathfrak{m} \in \mathrm{Spm}(R)} k) \otimes_k S \hookrightarrow \prod_{\mathfrak{m} \in \mathrm{Spm}(R)} (k \otimes_k S) = \prod_{\mathfrak{m} \in \mathrm{Spm}(R)} S.$$ Therefore $R \otimes_k S$ is a subring of a product of reduced algebras, therefore also reduced. $~\square$

If $k$ is not algebraically closed, the Lemma fails, even for fields. In fact, for a polynomial $f \in k[x]$ with splitting field $L$ the tensor product $k[x]/(f) \otimes_k L$ is isomorphic to the product of the algebras $L[x]/(x-\alpha)^{v_{\alpha}}$, where $\alpha$ runs through the roots of $f$ and $v_\alpha$ is its multiplicity. This algebra is reduced iff $v_\alpha=1$ for all $\alpha$ iff $f$ is separable. For example, $\mathbb{F}_p(t) \otimes_{\mathbb{F}_p(t^p)} \mathbb{F}_p(t) = \mathbb{F}_p(t)[x]/(x-t)^p$ is not reduced. The Lemma also fails when $k$ has characteristic zero, but then there is no counterexample for fields.

By the way, the isomorphism $k[A \times B] \cong k[A] \otimes_k k[B]$ holds almost by definition for affine schemes $A,B$. In this context the Lemma translates to the statement that the product of two reduced $k$-schemes is again reduced.

Answer 2

Let us assume that $K$ is algebraically closed.

If we have two $K$-algebras $C$ and $D$, there are canonical morphisms $$ C\overset{\alpha}{\longrightarrow} C\otimes_KD\overset{\beta}{\longleftarrow} D $$ defined by $\alpha(c)=c\otimes 1_D$ and $\beta(d)=1_C\otimes d$. If you are given two morphisms of $K$-algebras $\phi:C\to E$ and $\psi:D\to E$ to a third $K$-algebra $E$, then the universal property of tensor product (in the category of $K$-algebras) tells you that there exists a unique morphism of $K$-algebras $q:C\otimes_KD\to E$ such that $\phi=q\circ\alpha$ and $\psi=q\circ\beta$. In other words, in the category of $K$-algebras, $C\otimes_KD$ together with the arrows $\alpha$ and $\beta$ satisfies the universal property above.

Now, there is a duality of categories between the category of $K$-algebras of finite type and the category of affine varieties. This duality is given by the functor Spec. So take $$ \textrm{Spec (universal property of } C\otimes_KD),$$ after having assumed that $C$ and $D$ are of finite type (hence of the kind $C:=K[A]$ and $D:=K[B]$ as in your question). The universal property becomes a new universal property in the category of affine varieties, and now the universal object is $\textrm{Spec}(C\otimes_KD)$ with the two projections (the images $\textrm{Spec}(\alpha)$ and $\textrm{Spec}(\beta)$). But this is the universal property of $\textrm{Spec }C\times_K\textrm{Spec }D$, so that $$ \textrm{Spec}(C\otimes_KD)\cong \textrm{Spec }C\times_K\textrm{Spec }D=A\times_KB. $$ Now, to recover $K[A\times_KB]\cong K[A]\otimes_KK[B]=:C\otimes_KD$ it is enough to go through the duality in the other direction: since the inverse of Spec is $K[-]$ (the functor "global sections", or "take the coordinate ring of"), just take global sections of the last displayed formula to get what you want.

[Note that the identity you wrote between the ideals follows now as an obvious corollary.]

Answer 3

At the request of Martin Brandenburg, here is the result which may help you:

Let $k$ be a field and $V\subseteq \Bbb{A}_k^n$, $W\subseteq \Bbb{A}_k^m$ affine algebraic sets ( the proof I believe will also work if $V,W$ are affine varieties). Then $$I(V \times W )= I(V) + I(W).$$ By $I(V)$ we mean now the extension of the ideal $I(V)$ in the polynomial ring $k[x_1,\ldots,x_{m+n}]$.

Proof: See the discussion in my question here.

Using this result we can now prove the following theorem.

Theorem (Coordinate ring of a product): Let $V,W$ be as before. Then as $k$ - algebras we have $$k[V \times W] \cong k[V] \otimes_k k[W].$$

Proof: Let us write $I = \mathcal{I}(V)$ and $J = \mathcal{I}(W)$ and define $R = k[x_1,\ldots,x_n]$ and $S = k[x_{n+1},\ldots,x_{m+n}]$. Then by the lemma above, we have that $\mathcal{I}(V \times W) = I^e + J^e$ where the superscript denotes ideal extension in the ring $T = R \otimes_k S$, the polynomial algebra in $m+n$ variables. Now by the usual extension of scalars process we have $$I^e + J^e = I \otimes_k S +R \otimes_k J.$$ Thus to prove the theorem we need to prove that

$$\frac{T}{I \otimes_k S +R \otimes_k J} \cong \frac{R}{I} \otimes_k \frac{S}{J}.$$

The only tricky part in the proof is to construct a well-defined $k$ - algebra homomorphism $$f : \frac{R}{I} \times \frac{S}{J} \to \frac{ T}{I \otimes_k S + R \otimes_k J}$$

We can define one by declaring that $f$ sends $(\bar{a},\bar{b}) $ to $a\otimes b \mod{(I \otimes_k S + R \otimes_k J)}$. Is this well defined? Suppose $(\bar{a},\bar{b}) = (\bar{c},\bar{d})$. Then notice that $$\begin{eqnarray*} a \otimes b - c\otimes d &=& a \otimes b + c\otimes b - c\otimes b - c\otimes d \\ &=& (a-c) \otimes b + c \otimes (b-d)\\ &=& 0 \mod{(I \otimes_k S+ R \otimes_k J)}\end{eqnarray*}$$ because $a -c \in I$ and $b-d \in J$. Thus we see that $f$ is well-defined and leave the rest of the details for your to fill in, including constructing an inverse map. If you have any questions, I can add more details.